Let $k \in \mathbb{N}$. Show that there exist constants $C_1$ and $C_2$ so that
$$C_1(1+||x||^2)^k \leq \sum_{|\alpha| \leq k} x^{2\alpha} \leq C_2 (1+||x||^2)^k,$$
for all $x \in \mathbb{R}^n$. Remember that $x^{\alpha}=x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$. I am trying to use the inequality $$2|ab|\leq a^2+b^2$$. But the question sugest use some teory of tempered distribution.
The guess in the comments ( $(1+\|x\|^2)^k = \sum_{|\alpha|\le k} x^{2\alpha}$) is not correct; since none of the indices repeat, all coefficients on the right are +1, while the left hand side has positive coefficients, at least as large as binomial coefficients. Alternatively, verify that the identity is wrong if $x=(1,0,\dots,0)$: the right hand side is $k+1$ while the left hand side is $2^k$.
For the upper bound, $$\sum_{|\alpha|\le k} x^{2\alpha} \le \sum_{j=0}^k \|x\|^{2j} \sum_{|\alpha|=j}1 \le C \sum_{j=0}^k \binom{k}{j}(\|x\|^2)^j = C (1+\|x\|^2)^k. $$ Here, $C = \sup_{j=0}^k\sum_{|\alpha|=j}1$ is a constant depending on the known quantities $k,n$.
For the lower bound, if $j\neq 0$, then $$ \sum_{|\alpha|=j} x^{2\alpha} = \|x\|^{2j} \sum_{|\alpha| = j} \frac{x^{2\alpha}}{\|x\|^{2j}} \ge \|x\|^{2j} \frac{\sum_{m=1}^n x_m^{2j}}{\|x\|^{2j}} \ge c\|x\|^{2j}$$ where $c = \inf_{x\in\mathbb R^n} f(x)$ where $f(x) := \frac{\sum_{m=1}^n x_m^{2j}}{\|x\|^{2j}} \ge 0$. Note that $c=d(j)^{2j}>0$ by the equivalence of the $\ell^2$ and $\ell^{2j}$ norms on $\mathbb R^n$. ($d(j)$ depends on $n$ and $j$.) A similar bound obviously holds for $j=0$. Now $$\sum_{|\alpha|\le k} x^{2\alpha} \ge \frac{\inf_{j'=0}^k d(j')^{2j'}}{\max_{j''=0}^k \binom{k }{j''}}\sum_{j=0}^k \binom{k }{j}(\|x\|^2)^j = \tilde c (1+\|x\|^2)^k. $$
PS this question doesn't have anything to do with the
fractional-sobolev-spacestag. (and your other questions too)